10.2.3 Perimeter and Area, PT3 Practice

Question 7:
In diagram below, AEC is a right-angled triangle with an area of 54 cm² and BCDF is a rectangle.
Calculate
(a) the perimeter, in cm, of the coloured region.
(b) the area, in cm², of the coloured region.


Solution:
$\begin{array}{l}\text{(a)}\\ \text{Given area of }△ACE\\ \frac{1}{2}\times AC\times 9=54\\ AC=54\times \frac{2}{9}\\ AC=12\text{ cm}\\ \\ \text{Using Pythagoras’ theorem:}\\ AE=\sqrt{9^2+{12}^2}\\ =15\text{ cm}\\ \\ \text{Perimeter of coloured region}\\ =6+4.5+6+4.5+15\\ =36\text{ cm}\end{array}$

$\begin{array}{l}\text{(b)}\\ \text{Area of the coloured region}\\ =\text{Area of }△ACE-\text{Area of rectangle }BCDF\\ =54-\left(6\times 4.5\right)\\ =54-27\\ =27{\text{ cm}}^2\end{array}$

Question 8:
Diagram below shows a trapezium ABCDE. ABGF is a square with an area of 36 cm².


Calculate
(a) the perimeter, in cm, of the coloured region.
(b) the area, in cm², of the coloured region.


Solution:


$\begin{array}{l}\text{(a)}\\ \text{Using Pythagoras’ theorem:}\\ \text{In }△CDH,\\ CD=\sqrt{8^2+6^2}\\ =10\text{ cm}\\ \\ AB=BG=GF=FA=\sqrt{36}=6\text{ cm}\\ \\ \text{Perimeter of coloured region}\\ =6+10+18+2+6+6\\ =48\text{ cm}\end{array}$

$\begin{array}{l}\text{(b)}\\ \text{Area of the coloured region}\\ =\text{Area of trapezium }ABCDE-\text{Area of square }ABGF\\ =\left[\frac{1}{2}\left(12+18\right)\times 8\right]-36\\ =\left[\frac{1}{2}\times 30\times 8\right]-36\\ =120-36\\ =84{\text{ cm}}^2\end{array}$